Question

我对PHP比较陌生，需要一些帮助检查才能看到1.如果图像已上传提交并完成一组提交ELSE如果图像尚未上传，请提交一组不同的sql信息。

我之前尝试过isset（），is_uploaded_file（）和file_exists（），似乎无法发送任何内容。

HTML表格

           <form method="POST" action="php/editInv.php" enctype="multipart/form-data" id="editInv">
           <table>
              <tbody>
              <?php
                 $productId = $_GET['productId'];

                 require ('php/dbcon.php');

                 $con=mysqli_connect(HOST,USER,PASS,DB);

                 $sqlId = "SELECT * FROM products WHERE productId='".$productId."'";

                 $sqlIdResult = mysqli_query($con,$sqlId);

                 while($rowId=mysqli_fetch_array($sqlIdResult)) {
              ?>
                 <tr>
                    <td>
                       <label for="productName">Product Name: </label>
                    </td>
                    <td>
                       <input name="productName" id="productName" type="text" value="<?php echo $rowId['productName']; ?>"/>
                       <input name="prodId" id="prodId" type="hidden" value="<?php echo $rowId['productId']; ?>" />
                    </td>
                    <td>
                       <label for="productDesc">Product Description: </label>
                    </td>
                    <td rowspan="4">
                       <textarea name="productDesc" id="productDesc" cols="55" rows="10"><?php echo $rowId['productDesc']; ?></textarea>
                    </td>
                 </tr>
                 <tr>
                    <td>
                       <label for="gender">Gender: </label>
                    </td>
                    <td>
                       <select name="gender" id="gender">
                          <option id="1" value="1">Male</option>
                          <option id="2" value="2">Female</option>
                          <option id="3" value="3">Unisex</option>
                       </select>
                       <input type="hidden" value="<?php echo $rowId['genderId']; ?>" id="genderHid"/>
                       <script>
                          var gender=$('#genderHid').val();
                          var selected=$('#gender').find('#'+gender);
                          $(selected).attr('selected','selected');
                       </script>
                    </td>
                 </tr>
                 <tr>
                    <td>
                       <label for="inventory">Inventory: </label>
                    </td>
                    <td>
                       <input type="number" name="inventory" id="inventory" value="<?php echo $rowId['inventory']; ?>"/>
                    </td>
                 </tr>
                 <tr>
                    <td>
                       <label for="price">Price: </label>
                    </td>
                    <td>
                       $<input type="text" name="price" id="price" value="<?php echo $rowId['price']; ?>" />
                    </td>
                 </tr>
                 <tr>
                    <td>
                       <label for="productImage">Upload Image: </label>
                    </td>
                    <td>
                       <input type="file" name="productImage" id="productImage" />
                    </td>
                    <td colspan="2">
                       <div id="progress" style="width:100%;">
                          <div id="bar" style="height:50px;background-color:blue;width:0%;">
                          </div>
                          <p id="percent"></p>
                       </div>
                    </td>
                 </tr>
                 <tr>
                    <td colspan="2" rowspan="2">
                       <img id="prevImage" style="" src="<?php echo $rowId['productImage']; ?>" />
                    </td>
                    <td id="response">

                    </td>
                    <td>
                       <button type="submit" id="editInv">Edit Item</button>
                    </td>
                 </tr>
                 <tr>
                 </tr>
              <?php
              };
              ?>
              </tbody>
           </table>
           </form>

AJAX

  var options = {
  beforeSubmit: function() {
     // pre submit callback
     $("#progress").show();
     $("#percent").html("0%");
  },
  data: {
     productName : $('#productName').val(),
     productDesc : $('#productDesc').val(),
     inventory : $('#inventory').val(),
     price : $('#price').val(),
     gender : $('#gender').val(),
     image : $('#prevImage').attr('src'),
     prodId : $('#prodId').val()
  },
  uploadProgress: function(event, position, total, percentComplete) {
     //during submission
     $("#bar").width(percentComplete+'%');
     $("#percent").html(percentComplete+'%');
  },
  success: function(msg) {
     //post submit call back
     $(".bar").css("width","100%");
     $(".percent").html('100%');
     $("#response").html(response.responseText);
  },
  complete: function(response) {
     if(response.responseText=="Invalid File"){
     } else {
        $("#response").html(response.responseText);
        //$("#addNew")[0].reset();
        //$("#prevImage").attr('src','').hide();
        $(".bar").css("width","0%");
        $(".percent").html('0%');
     }

  },
  error: function(response) {
     alert(response.responseText);
  }

};

$("#editInv").ajaxForm(options);

PHP

//If a file has been uploaded
if (!empty($_FILES["productImage"]["name"])) {

 $target_dir = $_SERVER['DOCUMENT_ROOT'] . "/images/inventory/";
 $target_file = $target_dir . basename($_FILES["productImage"]["name"]);
 $fileName = str_replace(' ', '', $productName);
 $target_file_insert = "/images/inventory/" . $fileName . ".jpg";
 $targetFileUpload = $target_dir . $fileName . ".jpg";


 $sql1 = "UPDATE products SET productName='".$productName."', productDesc='".$productDesc."', inventory='".$inventory."', price='".$price."', genderId='".$gender."', productImage='".$targetFileUpload."' WHERE productId='".$prodId."'";

 mysqli_query($con,$sql1);

 $uploadOk = 1;
 $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
 // Check if image file is a actual image or fake image
 if(isset($_POST["submit"])) {
     $check = getimagesize($_FILES["productImage"]["tmp_name"]);
     if($check !== false) {
         echo "File is an image - " . $check["mime"] . ".";
         $uploadOk = 1;
     } else {
         echo "File is not an image.";
         $uploadOk = 0;
     }
 }
 // Allow certain file formats
 if($imageFileType != "jpg" && $imageFileType != "jpeg" && $imageFileType != "JPG" && $imageFileType != "JPEG") {
     echo "Sorry, only JPG, and JPEG files are allowed.";
     $uploadOk = 0;
 }
 // Check if $uploadOk is set to 0 by an error
 if ($uploadOk == 0) {
     echo "Sorry, your file was not uploaded.";
 // if everything is ok, try to upload file
 } else {
     if (move_uploaded_file($_FILES["productImage"]["tmp_name"], $targetFileUpload)) {
         mysqli_close($con);
     } else {
         echo "Sorry, there was an error uploading your file.";
     }
 }
} else
//If a file has not been uploaded
/*if (empty($_FILES["productImage"]["name"])) */{
  $sql2="UPDATE products SET productName='".$productName."', productDesc='".$productDesc."', inventory='".$inventory."', price='".$price."', genderId='".$gender."', productImage='".$image."' WHERE productId='".$prodId."'";

  mysqli_query($con,$sql2);
}

Answer 1

您需要检查的路径是$targetFileUpload而不是$upload。

上传后检查：

if (!file_exists($targetFileUpload)){...}

Answer 2

好吧，它实际上是由两个不同的问题引起的，一个在PHP中，一个在JavaScript中调用变量的顺序。我必须使用页面顶部的jQuery作为全局变量调用除生成的图像之外的所有内容，然后使用表单动作调用将jQuery AJAX放在页面底部。

显然，使用jQuery调用页面底部字段的值会导致它们从后端发送填充页面的信息，而将它们放在页面顶部会导致它们中继将更改的输入数据更改为AJAX调用。

任何说明为什么会这样运作？

  <script>
      var productName = $('#productName').val();
      var productDesc = $('#productDesc').val();
      var inventory = $('#inventory').val();
      var price = $('#price').val();
      var gender = $('#gender').val();
      var prodId = $('#prodId').val();
  </script>

//所有HTML内容

<script>
  var options = {
  beforeSubmit: function() {
     // pre submit callback
     $("#progress").show();
     $("#percent").html("0%");
  },
  data: {
     productName : productName,
     productDesc : productDesc,
     inventory : inventory,
     price : price,
     gender : gender,
     image : $('#prevImage').attr('src'),
     prodId : prodId
  },
  uploadProgress: function(event, position, total, percentComplete) {
     //during submission
     $("#bar").width(percentComplete+'%');
     $("#percent").html(percentComplete+'%');
  },
  success: function(msg) {
     //post submit call back
     $(".bar").css("width","100%");
     $(".percent").html('100%');
     $("#response").html(response.responseText);
  },
  complete: function(response) {
     if(response.responseText=="Invalid File"){
     } else {
        $("#response").html(response.responseText);
        //$("#addNew")[0].reset();
        //$("#prevImage").attr('src','').hide();
        $(".bar").css("width","0%");
        $(".percent").html('0%');
     }

  },
  error: function(response) {
     alert(response.responseText);
  }

};


$("#editInv").ajaxForm(options);
  </script>

PHP if语句基本上如下所示：

if (!empty($_FILES["productImage"]["name"])) {
} else
//If a file has not been uploaded
if (empty($_FILES["productImage"]["name"])) {
}

感谢@ fred-ii-获取错误捕获器。

PHP - 检查上传的图像

2 个答案: